Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x + 10} = \dfrac{6x + 40}{x + 10}$
Explanation: Multiply both sides by $x + 10$ $ \dfrac{x^2}{x + 10} (x + 10) = \dfrac{6x + 40}{x + 10} (x + 10)$ $ x^2 = 6x + 40$ Subtract $6x + 40$ from both sides: $ x^2 - (6x + 40) = 6x + 40 - (6x + 40)$ $ x^2 - 6x - 40 = 0$ Factor the expression: $ (x - 10)(x + 4) = 0$ Therefore $x = 10$ or $x = -4$ The original expression is defined at $x = 10$ and $x = -4$, so there are no extraneous solutions.